Moo, I already tried that method and came to the same result.
I thought of it in a different way - the proof isn't at all rigorous; in fact it's just an insight not a proof.
By adding 3 to z we are shifting the point representing z (on the complex plane) 3 units to the right.
Now since z can be -3, this gives us |z+3|=0. So this has to be the minimum - the smallest possible value for a modulus being 0.
Also if z = 6, then |z+3| = 9, and you will find that whatever other point z you take, the modulus |z+3| will always be smaller than 9.
But as I said, this is only an insight. I can't get any further.
Do you see something that I don't?

Mar 15th 2008, 09:16 AM

Moo

Well, i need to think again about it. There must be a way to put a restriction to a and to make it fit with your supposition (which seems correct ! ;))

To show the minimum of $\displaystyle |z+3|$ is zero is to just note that $\displaystyle |z+3|\geq 0$. Now we claim this is the best lower bound, and therefore minimum. If $\displaystyle z=-3$ then $\displaystyle 1\leq |z|\leq 6$ and $\displaystyle |z+3| = 0$.